A surveyor measures the angle of elevation to the top of a building to be 70 degrees. The surveyor then walks 50 ft farther from
the base of the tower and measures the angle of elevation to be 50 degrees. The surveyor's angle-measuring device 5.5 ft from the ground. How tall is the building, to the nearest foot?
2 answers:
43.804 ft
Working;
Start by drawing the diagram as shown in the attached diagram;
Then use the following trigonometric relations;
Tan (50)=
Tan (70)=
You can then use the value of x to find the value of h. Then add 5.5 ft to h to get height of the building
Working;
Start by drawing the diagram as shown in the attached diagram;
Then use the following trigonometric relations;
Tan (50)=
Tan (70)=
You can then use the value of x to find the value of h. Then add 5.5 ft to h to get height of the building
Answer:
height of building,H = 5.5 ft + 105.2 ft = 111 ft to the nearest ft.
Step-by-step explanation:
in this question we have given angle of elevation=
when surveyor walks 50m farther from the base of the tower that time angle of elevation=
As shown in figure in triangle ABD
.............(1)
or h=2.74x
and in triangle ABC
...........(2)
put value of h in equation 2
we got,
therefore,
Height of the tower is,h=2.74\times 38.39ft
h=105.2ft
Therefore,
height of building,H = 5.5 ft + 105.2 ft = 111 ft to the nearest ft.
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Answer:
Step-by-step explanation:
The initial amount invested, P = $1,600
Danny's age at which the amount, <em>A</em>, is to be used for his college, t = 19 years
The number of equal annual payments, <em>m</em>, to be withdrawn from the amount = 4
The compound interest on the account, r = 7% = 0.07
Let, <em>A</em>, represent the amount at the year the annual withdrawals starts to be made, we have;
n = The number of times the interest is applied annually = 1
Therefore;
The amount, <em>m</em>, is withdrawn at start of Danny's first year in college to give the amount in the account = A - m
The amount in the account at the end of the first year with compound interest, r = (A - m)×(1 + r)¹ = (A - m)×(1 + r)
At the stat of the second year, the second withdrawal is made to give the starting amount = (A - m)×(1 + r) - m
The amount in the account at the end of the second year = ((A - m)×(1 + r) - m)×(1 + r)
At the start of the third year, the amount in the account = ((A - m)×(1 + r) - m)×(1 + r) - m
At the end of the third year, we have the amount in the account = (((A - m)×(1 + r) - m)×(1 + r) - m) × (1 + r)
At the start of the forth year, the last yearly installment is withdrawn from the account and we have 0 balance in the account.
Therefore, on the fourth year, we have the amount in the account = (((A - m)×(1 + r) - m)×(1 + r) - m) × (1 + r) - m = 0
(A - m)×(1 + r)³ - m×(1 + r)² - m × (1 + r) - m = 0
A×(1 + r)³ - m×(1 + r)³ - m×(1 + r)² - m × (1 + r) - m = 0
A×(1 + r)³ = m×(1 + r)³ + m×(1 + r)² + m×(1 + r) + m = m × ((1 + r)³ + (1 + r)² + (1 + r) + 1)
A×(1 + r)³ = m × ((1 + r)³ + (1 + r)² + (1 + r) + 1)
m = A×(1 + r)³/((1 + r)³ + (1 + r)² + (1 + r) + 1)
∴ m = 5,786.444×(1 + 0.07)³/((1 + 0.07)³ + (1 + 0.07)² + (1 + 0.07) + 1) ≈ 1,596.56
The amount withdrawn annually, m ≈ $1,596.56
The amount in the account at the end of the each year is given as follows;
First year = (A - m)×(1 + r) = (5,786.444 - 1,596.56)×(1 + 0.07) ≈ 4,483.17588
Second year = ((A - m)×(1 + r) - m)×(1 + r) = ((5,786.444 - 1,596.56)×(1 + 0.07) - 1,596.56)×(1 + 0.07) = 3,088.68
Third year = (((A - m)×(1 + r) - m)×(1 + r) - m) × (1 + r) = (((5,786.444 - 1,596.56)×(1 + 0.07) - 1,596.56)×(1 + 0.07) - 1,596.56) × (1 + 0.07) = 1,596.57
At the end of the first year, we have $4,483.17588
At the end of the second year, we have $3,088.68
At the end of the third year, we have $1,596.57
At the end of the fourth year, we have 0
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r is the radius so plug that in.
if you only have the diameter, just divide that number by 2 to get the radius